Series T is a sequence of numbers where each term after the

Bravo, I found my mistake thanks to you. I was obsessing about "x times greater ...". But I solved it. So lets pretend the stem says ... x times greater ... Can someone give a solution to approach this?

You won't get the answer in this case, as you can not calculate: \(a+ax+ax^2+ax^{n-3}+ax^{n-2}+ax^{n-1}\) just knowing that \(a+ax^{n-1}=14\)

This is the way i solved it ..


The question gives us a general formula that confirms that we are dealing with an A.P. , therefore we know that the nth term = a (n-1)d .. In this case d = x ...

First term of the AP = A ; Last term = L ..

we know that A + L = 14 ...........[I]

Now we are to find the sum of 6 terms , ie the first three and the last three ... Lets assume that this series has 6 numbers .. therefore Sum of 6 numbers can be written as

S6 = 6/2 [ A + L ]

= 3 x 14 ( From I)

= 42 (D)


As an alternative to this method , we could PLUG IN values for X and solve for the sum of 6 terms of the AP as follows ------------->

Let us assume x =1 , therefore we would need to come up with a sequence whose first and last term adds to 14 , and which as at least 6 terms ...

let us use this sequence ...

3 , 4 , 5, 6, 7, 8, 9, 10, 11

The first and the last terms add to 14 ...

Each term is x greater than the one it succeeds ( in this case i have assumed x to be 01)

The first three and the last three terms are : 3 + 4 + 5 + 9 + 10 + 11 = 42 (D)


To further test the plug in method (not required to confirm the answer but just to test it out) ...

Now let us assume that x = 2

1 3 5 7 9 11 13

First three and last three terms add up to = 42 (D) ....

this is the way i solved it though it is long and took me over 5 mins. I guess Bunnel's method is best. Here is mine:

series is k, k+x, k+2x.......k+(n-1)x
sum of first and the last term is therefore k + k+(n-1)x = 14

summing first three members of the set, we have: k + k+x + k+2x = 3(k+x)--------(1)

summing the last 3 members of the set we have: k+(n-1)x + k+(n-2)x + k+(n-3)x = 3k+(3n-6)x ------(2)

(1)+(2) we have : 3[2k+(n-1)x]
the item in [] was given as 14

hence answer is 3*[14] = 42 .....(D)

Since no information is given about the first term, x, and anything else,
assume first term = 1; hence last term = 13,
so, 1 3 5 7 9 11 13 will suffice the series in which x = 2, from here answer is 42
also, 1 2 3 4 5 6 7 8 9 10 11 12 13 also suffices, where x = 1, from which the answer is again 42
bingo!!
(Saves time)

a1+a1+x+a1+2x+an+an-x+an-2x=3(a1+an)=3*14=42

if first three terms and last three terms are to sum differently,
then sequence needs a minimum of four terms
assume four terms: t1, t1+x, t1+2x, t1+3x
t1+(t1+3x)=14→
2*t1+3x=14
assuming t1 and x are positive integers,
then t1=4 and x=2
sequence=4, 6, 8, 10
sum of first 3 terms=18
sum of last 3 terms=24
combined sum=42
D

Hello Bunuel

Could you please tell me if my reasoning is good here?

Avg = 7

7 x 6 terms = 42

Because I am seeing that everybody is just multiplying 14 x 3 so I think I am doing something wrong.

Kind regards!

let first term be a, then second term is a+x, third term is a+2x, .....last but third term is a+(n-2)x, last but second a+(n-1)x and last term is a+nx

sum of first and last term, a+a+nx = 2a+nx = 14.

now as per q, sum of first three terms and last three terms = 6a+3nx = 3(2a+nx) = 3(14) = 42.

hence, D is corrrect answer

Am I mistaken to think that we can't solve this question without knowing (assuming) if X is an integer or Number of Terms? We could derive X by knowing what is A1 but as written the progression can start with a negative or 0. Can someone explain where I am flawed pls. Many thanks!

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